Solve Centripetal Force for m_1 in Frictionless Table

[klm]

are you saying that m2= m1 v^2/r

 

[klm]

so m2g= m1 v^2/r

 

[Kurdt]

are you saying that m2= m1 v^2/r

Almost. Remember weight is the mass multiplied by the acceleration due to gravity.

W_2 = m_2g

Can you work out v from there?

 

[klm]

square root (m2xgxr/ m1) = v ?

 

[Kurdt]

square root (m2xgxr/ m1) = v ?

Yes. Sorry I wasn't clearer earlier.

 

[klm]

no i am sorry. i have a hard time understanding physics, and i really appreciate the help you have given me.

 

[Kurdt]

So do you understand how this problem works now more importantly?

 

[klm]

i think so, though i do have one question. how come normal force and weight canceled each other out in m1? does that always happen in centripetal force?

 

[Kurdt]

Well m₁ is resting on a table which means its weight is counter acted by a normal force. That just comes from Newton's 3rd law of equal and opposite reactions. Also the weight and the normal force on m₁ act perpendicular to the centripetal force so they don't really come into the problem.

If I'm interpreting you wrong and you're asking why the centripetal force canceled the weight then simply because that's what the question asked you to do. So you set both equations equal and found out what speed m₁ would have to travel in a circle to balance m₂. Of courseif you were to set this up in a lab and you randomly spun the mass on the table it might be going too fast or too slow to balance the weight of m₂.

If that doesn't answer it then I'm not sure what weight you're referring to and what normal force.

 

[saket]

i think so, though i do have one question. how come normal force and weight canceled each other out in m1? does that always happen in centripetal force?

No, it is NOT a rule of thumb.
I suggest you making a free body diagram (fbd) of m_1. You already have identified the forces acting on it, when seen from ground, as the weight, normal force and the tension. (ur first thread.)
In the fbd of m_1 u will see that, in this problem, in the vertical direction only weight and normal force are acting. But since m_1 is neither lifting up from the table nor breaking into the table, (note that m_1 has to move in the plane of table), acceleration of m_1 in vertical direction is zero. Using Newton's 2nd Law: m_1.g - N = m_1.(0) => N = m_1.g
We left out tension, T, in so far discussion of fbd because it was in a perpendicular to the above two forces and as u very well be knowing that component in a perpendicular direction is FCos90 = 0.
Now in the horizontal direction (ie, along the plane of table), only force is this tension. Therefore, this must provide necessary centripetal force for the m_1 to move in a circular path.

 

[saket]

saket please don't take this personally but I do not think you're helping much.

it's okay sir. i m just a new entry to this forum.. i m still learning how to handle things. go on.. u r recognized homework-helper.

 

[klm]

thank you both very much. i understand now about the forces canceling out.